If a = lim_{n rightarrow infty} frac{1+2+3+ld | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) First,we evaluate $a$: $a = \lim_{n \rightarrow \infty} \frac{n(n+1)}{2n^2} = \lim_{n \rightarrow \infty} \frac{n^2+n}{2n^2} = \lim_{n \rightarrow

Vedclass · Accepted Answer

$2a = 3b$

Vedclass · Answer

(B) First,we evaluate $a$: $a = \lim_{n \rightarrow \infty} \frac{n(n+1)}{2n^2} = \lim_{n \rightarrow \infty} \frac{n^2+n}{2n^2} = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1}{2}$. Next,we evaluate $b$: $b = \lim_{n \rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3} = \lim_{n \rightarrow \infty} \frac{2n^3 + 3n^2 + n}{6n^3} = \lim_{n \rightarrow \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} = \frac{2}{6} = \frac{1}{3}$. Comparing the values,$a = \frac{1}{2}$ and $b = \frac{1}{3}$. Therefore,$2a = 2(\frac{1}{2}) = 1$ and $3b = 3(\frac{1}{3}) = 1$. Thus,$2a = 3b$.

If $a = \lim_{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ and $b = \lim_{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$,then

Similar Questions

If $f(x) = \begin{cases} x & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ x^2 & \text{if } x > 0 \end{cases}$,then $\mathop {\lim }\limits_{x \to 0} f(x) = $

Evaluate: $\cos \left[ \lim_{x \rightarrow \infty} \frac{2 \pi |x| + \pi x}{|x| - 3x} + \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x \right)}{x^2} \right]$

Find the limit: $\mathop {\lim }\limits_{x \to 1} \frac{x^{2}+1}{x+100}$

The value of $\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$ is

If $\lim _{x \rightarrow 0}\left(\frac{1+cx}{1-cx}\right)^{1/x}=4$, then $\lim _{x \rightarrow 0}\left(\frac{1+2cx}{1-2cx}\right)^{1/x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module